poj 3159(差分约束经典题)

题目链接:http://poj.org/problem?id=3159思路:题目意思很简单,都与给定的条件dist[b]-dist[a]<=c,求dist[n]-dist[1]的最大值,显然这是差分约束的经典题,条件可以转化为dist[b]<=dist[a]+c,于是a->b直接连边,边权值为c,从而题目转化为图上求1->n的最短路,看了一下数据,30000个点,150000条边,果断用Dijkstra+priority_queue,1300MS+险过,orz.

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<queue>
 6 #include<vector>
 7 using namespace std;
 8 #define MAXN 30030
 9 #define inf 1<<30
10 typedef pair<int,int>Pair;
11 
12 struct Edge{
13     int v,w;
14     Edge(int vv,int ww):v(vv),w(ww){}
15 };
16 
17 int dist[MAXN];
18 bool mark[MAXN];
19 int n,m;
20 
21 vector<vector<Edge> >map;
22 
23 int Dijkstra(int u)
24 {
25     memset(mark,false,(n+2)*sizeof(bool));
26     for(int i=1;i<=n;i++)dist[i]=inf;
27     dist[u]=0;
28     priority_queue<Pair,vector<Pair>,greater<Pair> >Q;
29     Q.push(make_pair(0,u));
30     while(!Q.empty()){
31         Pair pp=Q.top();
32         Q.pop();
33         int dd=pp.first,u=pp.second;
34         if(mark[u])continue;
35         mark[u]=true;
36         for(int i=0;i<map[u].size();i++){
37             int v=map[u][i].v,w=map[u][i].w;
38             if(mark[v])continue;
39             if(dd+w<dist[v]){
40                 dist[v]=dd+w;
41                 Q.push(make_pair(dist[v],v));
42             }
43         }
44     }
45     return dist[n];
46 }
47 
48 int main()
49 {
50     int u,v,w;
51     scanf("%d%d",&n,&m);
52     map.clear();map.resize(n+2);
53     while(m--){
54         scanf("%d%d%d",&u,&v,&w);
55         map[u].push_back(Edge(v,w));
56     }
57     printf("%d\n",Dijkstra(1));
58     return 0;
59 }
60 
61 
62         
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posted @ 2013-07-23 11:19  ihge2k  阅读(203)  评论(0编辑  收藏  举报